Then, to open :
Pxx/i
f ——
6
Appendix VI.
and the opening stress/ ==
1093
And to close :
P =
closing stress being /
x
bti
__
bh
But these must balance :
bh ~ bJi
and x = -
so P cannot be placed further from the centre than -J ^, or the line
of pressure must be kept within the middle third of the voussoir
depth h.
The Masonry Arch, Fig. 966, is drawn in half only, the
loads being symmetrically disposed; and there are five voussoirs,
v v &c., to each half. The loads w <w &c, are found by
erecting perpendiculars a b c d ef, each load being the resultant
of the weight of wall between a pair of verticals combined with
the weight of the voussoir, the latter being, of course, propor-
tionately small. Set these loads out in the force diagram as
01, 12, 23, &c. By symmetry the force O o must be horizontal.
Take any pole Ox and draw the link polygon g h, then find r s
the resultant of Oo and 04 the pressures at skewback and key
respectively. Take points / and k and find a new polygon passing
through them, such that the link 0 o still remains horizontal and
meeting r s in /. Join / k and draw the line 4 02 || / k, thus giving
the new pole 02. Complete the polygon k /, which will indicate
the line of pressure or resistance between the voussoirs, the
magnitudes of the forces being found from the force diagram.
The line kl must lie within the middle third of the joint, as shewn
dotted, and if this does not occur everywhere a readjustment
must be made. The extreme possible conditions are shewn at
E and F, where the line m n is better than f q^ because it causes a
lower value for the pressures. The greatest pressure occurs at the
skewback or support, and the intensity must never be more than
the material will allow. If the line of resistance passes through
the centre the intensity is considered uniform, but if the line
touches *the dotted arc the maximum intensity will be twice
that of the mean.